# Show that $f_n \to f$ over $\| \cdot \|_\infty$ implies $f_n \to f$ over $\| \cdot \|_2$ and is $(C[a,b], \|\cdot\|_2)$ Banach?

Exercise :

Over the space $$C[a,b]$$ we define the norm $$\|f\|_2 = \sqrt{\int_a^bf(x)^2\mathrm{d}x}$$ (i) Show that if the sequence $$(f_n)$$ converges to $$f$$ with respect to $$\| \cdot \|_\infty$$, then it also converges to $$f$$ with respect to $$\| \cdot \|_2$$.

(ii) Is $$(C[a,b], \| \cdot \|_2)$$ a Banach space ?

Attempt :

(i) Let $$(f_n)$$ be a sequence defined over $$C[a,b]$$ that converges to $$f$$ with respect to the norm $$\| \cdot \|_\infty$$. Then, this means that $$\exists n_0 \in \mathbb N$$ :

$$\|f_n-f\|_\infty< e \Leftrightarrow \max_{x \in [a,b]}|f_n(x) - f(x)| < \epsilon \; \forall n \geq n_0$$

Now, let's consider the quantity $$\| f_n - f \|_2$$. This is defined as :

$$\|f_n - f\|_2 = \sqrt{\int_a^b (f_n-f)^2(x)\mathrm{d}x}=\sqrt{\int_a^b[f_n(x)-f(x)]^2\mathrm{d}x}$$

Now, the last expression can be rewritten as :

$$\sqrt{\int_a^b[f_n(x)-f(x)]^2\mathrm{d}x}=\sqrt{\int_a^b |f_n(x) - f(x)|^2\mathrm{d}x}$$

But, it is :

$$|f_n(x)-f(x)|<\max_{x \in [a,b]}|f_n(x)-f(x)| < \epsilon \; \forall n \geq n_0$$

Now, since $$|f_n(x)-f(x)| \geq 0$$ and $$\epsilon >0$$, we yield :

$$|f_n(x)-f(x)|<\epsilon \Rightarrow |f_n(x)-f(x)|^2 < \epsilon^2 \equiv \epsilon' \; \forall n\geq n_0$$

And by integrating from $$a$$ to $$b$$ since $$f_n, f$$ are defined over $$C[a,b]$$ :

$$\int_a^b|f_n(x)-f(x)|^2\mathrm{d}x < \int_a^b\epsilon'\mathrm{d}x=\epsilon'(b-a) \equiv \epsilon'' \forall n\geq n_0$$

Thus, we have $$\|f_n-f\|_2 < \epsilon'' \; \forall n\geq n_0$$, which means that $$(f_n)$$ converges to $$f$$ with respect to the $$\|\cdot\|_2$$ norm.

Question : Is my approach to (i) correct ? How would one approach (ii) though ?

Part (i) looks fine, for the second part, define the sequence, $$f_n(x)= \begin{cases} 0,& a\leq x\leq a+\frac{b-a}{2}\\ \frac{2n}{(b-a)}(x-a-\frac{b-a}{2}),& a+\frac{b-a}{2} for which it is definitely better to draw a picture (we are taking continuous ramps between $$0$$ and $$1$$, with steeper slope, giving the ramp less and less time to get to $$1$$ as $$n$$ gets large). The point is that $$f_n(x)$$ is approximating the discontinuous function $$\chi_{[\frac{b-a}{2},1]}(x)$$ in $$||\cdot||_2$$, since $$||f_n(x)-\chi_{[\frac{b-a}{2},1]}(x)||\leq \frac{b-a}{4n}$$ where the bound comes from the area of the triangle between the two graphs (everything is $$\leq 1$$, so squaring only helps us). Since each $$f_n$$ is continuous and has a discontinuous limit in this metric, $$(C([a,b]),||\cdot||_2)$$ is not a Banach space.
• I have tried to add some intuition in the post, but the point is to try and get close in this norm to the simplest discontinuous function with a continuous one. I recommend drawing a picture and then using precalculus to figure out what the $f_n$ in the picture are – operatorerror Oct 28 '18 at 19:59
The approach to part (i) is correct. As for part (ii), Consider the sequence of functions $$f_{n}(x)=I_{[a,a+\frac{1}{n}]}(x)$$ .It converges in $$L^2$$, but is this sequence uniformly convergent?
• This isn't an answer to the question. The question is not whether uniform convergence implies onvergence in $L^2$ – operatorerror Oct 28 '18 at 19:58